There are n straight lines in a plane,no two | vedclass

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(C) Since no two lines are parallel and no three are concurrent,the $n$ straight lines intersect at $N = ^nC_2 = \frac{n(n-1)}{2}$ points.To form a li

Vedclass · Accepted Answer

$\frac{n(n - 1)(n - 2)(n - 3)}{8}$

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(C) Since no two lines are parallel and no three are concurrent,the $n$ straight lines intersect at $N = ^nC_2 = \frac{n(n-1)}{2}$ points.To form a line,we need $2$ points. The total number of lines formed by joining these $N$ points is $^NC_2$.However,this total includes the original $n$ lines. Each original line contains $(n-1)$ intersection points. The number of lines formed by these $(n-1)$ points on a single original line is $^{n-1}C_2$. Since there are $n$ such lines,we must subtract the lines that are part of the original set.The number of fresh lines is $^NC_2 - n 	imes ^{n-1}C_2$.Substituting $N = \frac{n(n-1)}{2}$:$= \frac{N(N-1)}{2} - n \frac{(n-1)(n-2)}{2}$$= \frac{\frac{n(n-1)}{2} (\frac{n(n-1)}{2} - 1)}{2} - \frac{n(n-1)(n-2)}{2}$$= \frac{n(n-1)(n^2-n-2)}{8} - \frac{4n(n-1)(n-2)}{8}$$= \frac{n(n-1)(n-2)(n+1) - 4n(n-1)(n-2)}{8}$$= \frac{n(n-1)(n-2)(n+1-4)}{8} = \frac{n(n-1)(n-2)(n-3)}{8}$.

There are $n$ straight lines in a plane,no two of which are parallel and no three pass through the same point. Their points of intersection are joined. Then the number of fresh lines thus obtained is

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